Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $y = \dfrac{p^2 - 12p + 36}{-8p + 56} \div \dfrac{3p^2 - 18p}{p - 7} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{p^2 - 12p + 36}{-8p + 56} \times \dfrac{p - 7}{3p^2 - 18p} $ First factor the quadratic. $y = \dfrac{(p - 6)(p - 6)}{-8p + 56} \times \dfrac{p - 7}{3p^2 - 18p} $ Then factor out any other terms. $y = \dfrac{(p - 6)(p - 6)}{-8(p - 7)} \times \dfrac{p - 7}{3p(p - 6)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (p - 6)(p - 6) \times (p - 7) } { -8(p - 7) \times 3p(p - 6) } $ $y = \dfrac{ (p - 6)(p - 6)(p - 7)}{ -24p(p - 7)(p - 6)} $ Notice that $(p - 7)$ and $(p - 6)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(p - 6)}(p - 6)(p - 7)}{ -24p(p - 7)\cancel{(p - 6)}} $ We are dividing by $p - 6$ , so $p - 6 \neq 0$ Therefore, $p \neq 6$ $y = \dfrac{ \cancel{(p - 6)}(p - 6)\cancel{(p - 7)}}{ -24p\cancel{(p - 7)}\cancel{(p - 6)}} $ We are dividing by $p - 7$ , so $p - 7 \neq 0$ Therefore, $p \neq 7$ $y = \dfrac{p - 6}{-24p} $ $y = \dfrac{-(p - 6)}{24p} ; \space p \neq 6 ; \space p \neq 7 $